question_answer

In the given figure, ABC is a triangle and CHED is a rectangle.       \[BC=12cm,\] \[HF=6cm,\] \[\text{FC}=\text{BF}\]and altitude AF is 24 cm. The area of rectangle is:                 

A) \[56c{{m}^{2}}\]

B) \[54c{{m}^{2}}\]

C) \[60c{{m}^{2}}\]

D) \[72c{{m}^{2}}\]

Correct Answer: B

Solution :

[b] In \[\Delta AFC,\] \[AF||HE\]

\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\Delta AFC\tilde{\ }\Delta EHC\]

\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\frac{AF}{HE}=\frac{FC}{HC}\]

\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\frac{24}{6}=\frac{6}{HC}\]

\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,HC=3/2\,\,cm\]

\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,FH=6-3/2=9/2\,cm\]

Similarly, \[GF=9/2\]

\[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,GH=\frac{9}{2}+\frac{9}{2}=9\,cm\]

Area of rectangle \[GHED=GH\times HE\]

\[=9\times 6=54\,\,c{{m}^{2}}\]